الرياضيات المتناهية الأمثلة

$[\begin{array}{c} - 2 & 1 & - 3 \\ 0 & - 3 & 1 \end{array}] + [\begin{array}{c} x - y & 4 & 2 \\ - 4 & x & - 1 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 1

اجمع العناصر المتناظرة.

$[\begin{array}{c} - 2 + x - y & 1 + 4 & - 3 + 2 \\ 0 - 4 & - 3 + x & 1 - 1 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 2

Simplify each element.

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خطوة 2.1

أضف $1$ و $4$ .

$[\begin{array}{c} - 2 + x - y & 5 & - 3 + 2 \\ 0 - 4 & - 3 + x & 1 - 1 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 2.2

أضف $- 3$ و $2$ .

$[\begin{array}{c} - 2 + x - y & 5 & - 1 \\ 0 - 4 & - 3 + x & 1 - 1 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 2.3

اطرح $4$ من $0$ .

$[\begin{array}{c} - 2 + x - y & 5 & - 1 \\ - 4 & - 3 + x & 1 - 1 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 2.4

اطرح $1$ من $1$ .

$[\begin{array}{c} - 2 + x - y & 5 & - 1 \\ - 4 & - 3 + x & 0 \end{array}] = [\begin{array}{c} 0 & 5 & - 1 \\ - 4 & 2 x + y & 0 \end{array}]$

خطوة 3

Write as a linear system of equations.

$- 2 + x - y = 0$

$5 = 5$

$- 1 = - 1$

$- 4 = - 4$

$- 3 + x = 2 x + y$

$0 = 0$

خطوة 4

Since $0 = 0$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying the system.